Problem: $f(x, y) = \dfrac{x - 3}{y^2 + 1}$ What is the partial derivative of $f$ with respect to $y$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{y^2 + 1}$ (Choice B) B $\dfrac{-2y}{(y^2 + 1)^2}$ (Choice C) C $\dfrac{-2y(x - 3)}{(y^2+1)^2}$ (Choice D) D $\dfrac{y^2 + 1 - 2y(x - 3)}{(y^2 + 1)^2}$
Explanation: Taking a partial derivative with respect to $y$ means treating $x$ like a constant, then taking a normal derivative. We can use the chain rule to differentiate $(y^2 - 1)^{-1}$, noting that $(x - 3)$ acts like a constant here. $\begin{aligned} \dfrac{\partial f}{\partial y} &= \dfrac{\partial}{\partial y} \left[ \dfrac{x - 3}{{y^2} + 1} \right] \\ \\ &= (x - 3) \dfrac{\partial}{\partial y} \left[ ({y^2} + 1)^{-1} \right] \\ \\ &= (x - 3) \left[ -{2y}({y^2} + 1)^{-2} \right] \\ \\ &= \dfrac{-{2y}(x - 3)}{({y^2} + 1)^2} \end{aligned}$ In conclusion, $\dfrac{-2y(x - 3)}{(y^2 + 1)^2}$